[James_H._Harlow]_Electric_Power_Transformer_Engin(BookSee.org)

V
90
( X X )
s
T
(2.9.10a)
where, I SC = short circuit current, p.u.
Since (X S + X T ) is small, the short-circuit current (I SC ) can become very large. The total transmitted
power then equals the available power from the source (MVA SC ),
S = V S I SC (2.9.11)
FIGURE 2.9.28 Radial system.
V ( V)
s
S 90
( X X )
s
T
(2.9.11a)
where
V S = sending end voltage, p.u.
V R = receiving end voltage, p.u.
V = bus voltage drop, p.u.
I = bus current, p.u.
Z S = source impedance, p.u.
Z T = transformer impedance, p.u.
Z L = load impedance, p.u.
Under steady state, the load impedance 4 Z L essentially controls the current I, since both Z S and Z T are
small. Also, typically X S R S , X T X S , and the load P L Q L . Therefore,
V = V S – V R [PR S + Q(X S + X T )] (2.9.9)
where
P= real power, p.u.
Q = reactive power, p.u.
R S = source resistance, p.u.
X S = source reactance, p.u.
X T = transformer reactance, p.u.
V = voltage drop, p.u.
and V, per unit, is small
When a short circuit occurs (closing the switch SW), V R 0 and Z L = 0 (bolted fault), and Equation
2.9.8 can be rewritten as
I SC = V/[R S + j (X S + X T )] (2.9.10)
4
The load may, in fact, be a rotating machine, in which case the back electromotive force (EMF) generated by the
motor is the controlling factor in limiting the current. Nevertheless, this EMF may be related to an impedance, which
is essentially reactive in nature.
where, S = transmitted power, p.u.
Therefore, as voltage drops, the system voltage is shared between the system impedance (transmission
lines) and the transformer impedance:
V S = V Q S (X S + X T ) (2.9.12)
where, Q S = transmitted reactive power, p.u.
Since X T is typically much larger than X S , the voltage drop across the transformer almost equals the
system voltage. Two major concerns arise from this scenario:
1. Mechanical stresses in the transformer, with the windings experiencing a force proportional to
the square of the current
2. Ability of the circuit breaker to successfully interrupt the fault current
Therefore, it is imperative to limit the short-circuit current so that it will not exceed the ratings of
equipment exposed to it. Basic formulas are as follows:
Three-phase fault:
1-to-ground fault:
I 3 , kA = 100 [MVA]/[ 3 V LL Z 1 ] (2.9.13)
I SLG , kA = 3 100 [MVA]/[ 3 V LL Z T ] (2.9.14)
Z T = Z 1 + Z 2 + Z 0 + 3Z N
(2.9.14a)
where
V LL = line-to-line base voltage, kV
Z T = total equivalent system impedance seen from the fault, p.u.
Z 1 , Z 2 , and Z 0 = equivalent system positive-, negative-, and zero-sequence impedance
seen from the fault (in p.u. @ 100-MVA base)
Z N = any impedance intentionally connected to ground in the path of the fault current, p.u.
2.9.3.2 Phase Reactors vs. Bus-Tie Reactors
A method to evaluate the merits of using phase reactors vs. bus-tie reactors is presented here. Refer to
Figure 2.9.29 and Figure 2.9.30 and the following definitions:
I S1 and I S2 = available fault contribution from the sources, kA.
I L = rated current of each feeder, kA.
n = total number of feeders in the complete bus (assuming all feeders are identical)
I b = I S2 , the interrupting rating of the feeder circuit breakers, > 1, kA
© 2004 by CRC Press LLC
© 2004 by CRC Press LLC